Question 642955
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Depends.  Could be:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sqrt{32^2\ +\ 16^2}]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sqrt{32^2\ -\ 16^2}]


depending on whether the unknown side is the hypotenuse or not.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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